CS201 Week 1 Chapter1 Introduction

8/6/2019 CS201 Week 1 Chapter1 Introduction

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CS 201

Data Structures & Algorithms

Chapter 1 Introduction

Text: Read Weiss, 1.1 1.3

Izmir University of Economics


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Course Policy Syllabus

Grading

Labs always in C programming language

Each assignment starts and ends in the same

Lab session. Late assignments will not be

accepted. Study hard!



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3

Introduction

See that how a program performs for

reasonably large input is just as important as

its performance on moderate amounts of

input

Summarize basic mathematical background

needed Review recursion



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4

Motivating Examples: Selection

Selection problem: you have a group ofNnumbers and

would like to determine the kth largest.

I: read them into an array. Sort them in decreasing order.

Return the kth element.

II: read the first kelements into the array. Sort them indecreasing order. Next read the remaining elements one by

one. If the new element read is smaller than the last, ignore

it otherwise place in the correct spot in the array bumping

one element out of the array.

A simulation with a random file of 10 million elements andk = 5,000,000 shows that each requires several days of

computer processing.



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Motivating Examples: Word Puzzles

Solving a popular word puzzle: Input consists of a twodimensional array of letters and a list of words. Theobjective is to find the words lying horizontally, verticallyor diagonally in either direction.

I: for each word in the word list, check(row, column,

orientation) II: for each ordered quadruple (row, column, orientation,

number of characters), test whether the word is in the wordlist.

{this, two, fat, that}



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Math Review - Exponents

XAXB = XA+B

XA

/XB

= XA-B

(XA)B = XAB

XN+XN = 2XN !=X2N



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Math Review Logarithms I

In computer science, all logarithms are to the base 2 unlessspecified otherwise.

Definition 1.1. XA = B iff logXB=A

Theorem 1.1. logAB = logCB/logCAwhere A, B, C > 0, A != 1

Proof: Let X=logCB, Y=logCA, Z=logAB

CX=B, CY=A, AZ=B by Definition 1.1.

B=CX=(CY)Z.Therefore, X=YZ

Theorem 1.2. log AB = logA + logB where A, B > 0

Proof: X=logA, Y=logB, and Z=logAB,

2X=A, 2Y=B, and 2Z=AB, 2X2Y=AB=2Z.

Therefore, X+Y=Z



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Math Review Logarithms II

log A/B = logA logB

log(AB)=BlogA

logX < X for all X > 0

log1 = 0, log2 = 1, log1024 = 10

8Izmir University of Economics


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Math Review Series I Geometric Series

If 0 < A < 1, then and as N

tends to g, the sum approaches1/(1-A)

S=1+A+A2+A3+A4

... AS=A+A2+A3+A4+A5...

S-AS= 1 which implies S=1/(1-A)

112

0

2 !!

NN

i

i

1

1

0

1

!

!

A

AN

i

iA

N

A

N

i

iA

e

! 1

1

0



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Math Review Series II

Arithmetic Series

2...

2

1

2

1

2

112

...2

4

2

3

2

212

...2

4

2

3

2

2

2

1

?

1

2/

32

32

432

!!!

!

!

!

!

SSS

S

S

i

ii

2/)1(

1...1112

_______________________________

1...21

...321

)(2

2

2

)1(

1

!

!

!

!

}

!!

NNS

NNNNS

NNN

N

S

S

MethodGaussNNNN

i

i

2/)13(

2/)1(31

1

1

3

1

1

1

3

1

13

:xample

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!

!

!

kk

kkk

k

i

k

i

i

k

i

k

i

ik

i

i



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Math Review Series III

3/3

6/)12)(1(

6/)132

2(

6/)2)1(322(

3/)12/)1(32

(

3/)2/)1(33

(

2/)1(333

1 13

2

31 )13

2

3

3

(

3

3

1

3)1(

1

3

1

3)1(

3

1

2

N

NNNS

NNNS

NNNS

NNNS

NNNNS

NNNSN

N

i ii

N

i iiii

N

N

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11

1

1{

}!

kwhenk

kNN

iki

1log

1

1

SeriesHarmonicthN

!}!

! kwhenNe

N

i iN

H

The error in theapproximation tends to

Eulers constant

K = 0.57721566

!

!

!

!

!

!

10

1

)(

1

)(

0

)(

)(

1

)(

onsManipulatilgebraiceneral

n

i

ifN

i

ifN

ni

if

NNfN

i

Nf



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The P Word Proof by Induction There are various ways of proving statements in

data structures analysis

Proof by Induction: It has two standart parts: The

first step is proving a base case. Establishing that

a theorem is true for some small (usuallydegenerate) value(s). This step is almost always

trivial.

Next, an inductive hypothesis is assumed.

Generally this means that the theorem is assumedto be true for all cases up to some limit k.Using the

assumption, the theorem is then shown to be true

for the next value, typically k+1.



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The P Word Induction Example I Example: Prove that Fibonacci Numbers F0=1, F1=1, and

Fi=Fi-1+Fi-2 for i > 1, satisfy Fi < (5/3)i for i 1. Proof: Verify that the theorem is true for the trivial cases

(base cases): F1=1 < (5/3)1 and, F2 = 2 < (5/3)

2. These

prove the basis. We now assume that the theorem is true

for i = 1, 2, ..., k; this is the inductive hypothesis. To provethe theorem, we need to prove Fk+1


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Example: IfN 1 then

Proof: For the base case, the theorem is true when N= 1.For the inductive hypothesis, assume the theorem is true

for 1 k N. Lets try to prove that it is true forN+1

6/)12)(1(1

2 ! ! NNNNi

i

6

)32)(2)(1(6

6722)1(

)1(6

)12()1(

2)1(6

)12)(1(1

1

2

2)1(

1

21

1

2

!

!

!

!

!

!

!

!

NNN

NNN

NNN

N

HypothesisInductivebyNNNNN

i

i

NN

i

iN

i

i

The P Word Induction Example II



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Proof by Counterexample: Best way for

proving that a statement is false.

Example: The statement Fk k2 is false.

The easiest way to prove this is to compute

F11 = 144 > 112 = 121

16

Proof by Counterexample



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Proof by Contradiction

It proceeds by assuming that the theorem is false andshowing that this assumption implies that some known

property is false, and hence the original assumption is

erroneous.

Example: Prove that there is an infinite number of primes.

Proof: Assume the theorem is false, so that there is some

largest prime Pk. Let P1, P2, ..., Pkbe all the primes in

order and considerN= P1P2...Pk+ 1. Clearly, N> Pk, so by

assumptionNcan not be prime.

However, none ofP1, P2, ..., Pkdivides Nexactly, becauseremainders are all 1. This is a contradiction: numbers are

either prime or a product of primes. Hence the original

assumption is false implying that the theorem is true.



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A Brief Introduction to Recursion A function that is defined in terms of itself is called

recursive. Not all mathematically recursive functions arecorrectly or efficiently implemented by recursion.

Example: for all integers x 0 with f(0)=0

#include

int F( int X ) {

if( X == 0 ) /* base case */

return 0;

else

return 2 * F( X - 1 ) + X * X;

}

main( )

{

printf( "F(4) = %d\n", F( 4 ) );

return 0;

}

2)1(2)( xxfxf !

If F is called with a value of

4, then 2*F(3)+

4*4

will berequired to be computed. Thus

a call is made to find F(3).

F(4)=2*F(3)+4*4

F(3)=2*F(2)+3*3

F(2)=2*F(1)+2*2 F(1)=2*F(0)+1*1

F(0)=0 base case. Recursive

calls until a base case. F(-1)=

Automatic Bookkeeping



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Recursion - Bad

#include

int Bad( unsigned int N ) {

if( N == 0 )

return 0;

else

return Bad( N / 3 + 1 ) + N - 1;

}

main( )

{

printf( "Bad is infinite recursion\n" );

return 0;

}


Bad(0)=0, Bad(N)=Bad(N/3 + 1) + N 1

To compute Bad(1), the computer will repeatedly makecalls to Bad(1). Eventually, it will run out of space


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Fundamental Rules of Recursion - I 1) Base cases: You must always have some

base cases, which can be solved without

recursion.

2)Making progress: For the cases to be

solved recursively, the recursive call must

always be to a case that makes progresstoward a base case.

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3)D

esign rule: Assume that all therecursive calls work. This rule is important.

It relieves you of the burden of thinking

about the details of bookkeeping.

4) Compound interest rule: Never duplicate

work by solving the same instance of a

problem in separate calls.

Hidden bookkeeping costs are mostlyjustifiable. However; It should neverbe

used as a substitute for a simple loop.

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Fundamental Rules of Recursion - II

CS201 Week 1 Chapter1 Introduction

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